Exercise superposition difficulties

[esmeco]

Hello!

I'm having a bit of a difficulty trying to solve this exercise mostly because of the resistors.
So,when short circuiting the voltage source we have the 20ohm in parallel with 40 ohm right?But what about the other resistors are they in parallel or in series with the result 20//40?The exercise tells to determine the voltage with superposition in the 40ohm resistor.
And what about if we open circuit the current source?Are the 2.5 ohms and 10ohm in series with each other?
What kind of process should I adopt to solve for the voltage in 40ohm resistor?Currents in equal currents out?

Any help is really appreciated!

 

[esmeco]

Anyone can help me?

 

[Astronuc]

Short ciruciting the voltage source makes the 20Ω and 40Ω share two common nodes, so they are in parallel. That combination share one node with the 10Ω, so they are in series. Then that combination share two commons with the 2.5Ω resistor and the combination of resistances is in series with the current source.

And what about if we open circuit the current source?Are the 2.5 ohms and 10ohm in series with each other?

The 2.5Ω and 10Ω have one node in common. Therefore they are . . . ?

Voltage drop, V, across a resistor of resistance R is given by V = i R where i is the current through the resistor.

 

[ranger]

Try redrawing

One thing that always helps me is redrawing a messy looking circuit to something more conventional. This may help you to visualize what Astronuc was taking about. [Sorry for the sloppiness].

 

[esmeco]

But,doesn't the combination 20//40 also shares a node with 2.5ohm resistor besides the 10ohm?

 

[ranger]

The redrawn image has been approved. Following Astro's advice can you now make it out?

 

[Astronuc]

But,doesn't the combination 20//40 also shares a node with 2.5ohm resistor besides the 10ohm?

Yes, the 20//40 share one common node with the 2.5Ω resistor, and they share one common node with the 10Ω resistor, but those two common nodes are different.

The 20//40Ω + 10Ω are parallel with the 2.5Ω resistor.

 

[esmeco]

So,the KCl for the shorted voltage would be something like this:

va/2.5 + (va-vb)/10=6
(va-vb)/10=vb/10 + vb/20

Am I right?

Oh,and could the voltage in the 40 resistor in the open current source be calculated by 10 and 2.5 in series,in parallel with 40 volts,in series with 20 resistor,and then calculate the total current since we have the 100 voltage source.And with the total current calculated multiplied by the 40 ohm resistor in order to determine the tension in the resistor?

 

[huckmank]

Oh,and could the voltage in the 40 resistor in the open current source be calculated by 10 and 2.5 in series,in parallel with 40 volts,in series with 20 resistor,and then calculate the total current since we have the 100 voltage source.And with the total current calculated multiplied by the 40 ohm resistor in order to determine the tension in the resistor?

You were correct until the last sentence. The total current gets split between the 40 ohm branch and the 10 + 2.5 series branch. The best way to find that voltage is to keep the 40 || (10 + 2.5) combination together and use voltage divider for that combination and the 20 ohm. This works because the voltage across the 10 + 2 combination is the same as the voltage across the 40 since they're in parallel.

Voltage divider, if you don't recall is Vbranch = Vtotal * (Rbranch/Rtotal) and only works on series circuits.